Question

Let $S_k$ be the sum of an infinite GP series whose first term is $k$ and the common ratio is $\frac{k}{k+1}(k>0)$. Then, the value of $\sum _{k=1}^{\infty }\frac{{(-1)}^k}{S_k}$ is equal to

• A. ${\log }_e^4$
• B. ${\log }_e^2-1$
• C. $1-{\log }_e^2$
• D. $1-{\log }_e^4$

Answer 1

The correct option is D

$1-{\log }_e^4$

Determine the value of $\sum _{k=1}^{\infty }\frac{{(-1)}^k}{S_k}$

We know that sum of an infinite GP series is given by: $S=\frac{a}{1-r}$

$$\begin{gathered} S_k=\frac{k}{1-\left({\displaystyle \frac{k}{k+1}}\right)}\left[\because \text{common ratio}=\frac{k}{k+1},\text{First term}=k,\text{given}\right] \\ =\frac{k(k+1)}{k+1-k} \\ =k(k+1) \end{gathered}$$

Now,

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{{(-1)}^k}{S_k}=\frac{-1}{1.2}+\frac{1}{2.3}-\frac{1}{3.4}..........\infty \\ =-(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(-1)(\frac{1}{3}-\frac{1}{4})......\infty \\ =-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{3}-\frac{1}{3}.........\infty \\ =-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}.........\infty \right) \\ =-1-2\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}.........\infty \right) \\ =-1-2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}.........\infty \right)+2 \\ Since,{\log }_e^2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}.........\infty \\ \therefore 1-2{\log }_e^2\text{or}1-{\log }_e^4 \end{gathered}$$

Thus, the correct option is D.